Efficiency of Synchronous Machine

Efficiency of Synchronous Machine

The different possible losses in a synchronous machine are,

i) No load rotational losses which are further divided as (a) friction and windage losses and (b) open-circuit core losses

ii) Field winding copper losses (l²_fR_f)

iii) Armature winding copper losses (l²_aR_a) called direct load losses

iv) Stray load losses

The stray load losses and direct load losses together called short circuit load losses.

The open circuit core losses include eddy current and hysteresis losses which depend on the square of the open circuit voltage.    This   is shown in the Fig. 3.17.1 (a).         

These losses exist as long as   field winding is excited. The friction and windage losses are constant with respect to field current. These        losses exist though the field winding is not energised and machine   is running on no   load.  Thus the graph of rotational losses against field current has an intercept on vertical axis as shown in the Fig. 3.17.1 (b).

Short circuit load loss = [Stray load losses] + [Armature copper losses]

The variation of stray load losses against armature current is shown in the Fig. 3.17.2.

The stray load losses include core loss due to armature leakage flux and armature ohmic loss due to skin effect. The armature copper loss component is calculated by considering hot value of d.c. armature resistance generally at a temperature of 75°C.

The effective value of armature resistance is obtained as,

R_a (effective) = Stray load losses / (I_asc)²

where I_asc = Short circuit armature current. Once calculated at the rated current, Ra (effective) is constant at all load conditions.

The power flow diagram for three phase alternator is shown in the Fig. 3.17.3.

Once all the losses are known then the alternator efficiency can be obtained as,

% ƞ = output / output + Losses × 100

1. Maximum Efficiency

For a synchronous machine, Variable losses = Armature copper losses. Constant losses = Field copper losses + No load rotational losses

Similar to d.c. machine, the maximum efficiency occurs when variable losses are equal to the constant losses.

Derivation :

where I_a = I_am = Amature current at whcih maximum ƞ occurs.

Example 3.17.1 A 30 MVA, 15 kV, 60 Hz ac generator has a synchronous reactance of 1.2 pu and a resistance of 0.02 pu. Calculate. (1) the base voltage, base power and base impedance of the generator, (2) the actual value of the synchronous reactance, (3) the actual winding resistance, per phase (4) the total full-load copper losses. AU : May-17, Marks 8 Solution :

Solution :

2) X_S =  1.2 pu

X_S = X_S (pu) × Z_B = 1.2 × 7.5 = 92 Ω

3) R_a = 0.02 pu

R_a = R_a (pu) × Z_B = 0.02 × 7.5 = 0.15 Ω

4) On full load, per unit value of current is          unity i.e. 1 = 1 pu.

Per unit copper loss = I² × R_a (pu) = 0.02 pu (per ph)

Total actual copper loss = 3 × 0.02 pu × S_B

= 3 × 0.02 × 10 × 10⁶ = 600 kW

Examples for Practice

Example 3.17.2 A 500 kVA, 11 kV, 3ϕ star-connected, alternators has the following data :

Friction and windage loss = 1500 W

Open-circuit core loss     = 2500 W

Effective armature resistance/phase    = 4Ω

Field copper loss  = 1000 w

Find the following parts in regarding with above sychronous alternators :

i) Alternator efficiency of half-full load and at 0.85 power factor lagging.

ii) Maximum efficiency of the alternator.

[Ans.: i) %ƞ = 96.78 %, ii) %^m = 97.06 %]

Example 3.17.3 A 3 ϕ, hydro-electric synchronous generator is read to be 110 MW, 0.8 pf. lagging, 6 kV, Y-connected, 50 Hz, 100 rpm. Determine the following

i) The number of poles, ii) The kVA rating iii) The prime-mover rating if the full-load generator efficiency is 97.1 % (level out field loss)

[Ans.: i) 60 poles ii) 137.5 x103 kVA iii) 113.285 MW]

Review Question

1. Write a note on losses in a synchronous machine.